Prove that $6<3^\sqrt3$ implies $3^\sqrt3 < 7$

Claim: $7 > 3^{\sqrt 3}$. Otherwise $$7^\sqrt 3 \le 27 < 28 = 7*4 \\ 7^{\sqrt 3 - 1} < 4 \\ 48 < 7^2 < 4^{\sqrt 3 + 1} \text{(since }(7^{\sqrt 3 - 1})^{\sqrt 3 + 1} = 7^{(\sqrt 3 - 1)(\sqrt 3 + 1)} = 7^{(\sqrt 3)^2 - 1^2} = 7^2{)} \\ 3 < 4^{\sqrt 3 - 1} \: \text{(divide by 16)}\\ 3^{\sqrt 3 + 1} < 4^2 = 16 < 18 \\ 3^\sqrt 3 < 6$$ ...and we have a contradiction with a given statement.

$$6=(6^3)^{\frac{1}{3}}=216^{\frac{1}{3}}<243^{\frac{1}{3}}=(3^5)^{\frac{1}{3}}=3^{\frac{5}{3}}<3^{\sqrt{3}},$$ becouse $$\left(\frac{5}{3}\right)^2=\frac{25}{9}<3=\left(\sqrt{3}\right)^2.$$ Since $$\left(\sqrt{3}\right)^2=3=\frac{48}{16}<\frac{49}{16},$$ then $$\sqrt{3}<\frac{7}{4}.$$

Therefore $$3^{\sqrt{3}}<3^{\frac{7}{4}}=(3^7)^\frac{1}{4}=2187^{\frac{1}{4}}<2401^{\frac{1}{4}}=(7^4)^{\frac{1}{4}}=7.$$